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1/2r+6=3-2r
We move all terms to the left:
1/2r+6-(3-2r)=0
Domain of the equation: 2r!=0We add all the numbers together, and all the variables
r!=0/2
r!=0
r∈R
1/2r-(-2r+3)+6=0
We get rid of parentheses
1/2r+2r-3+6=0
We multiply all the terms by the denominator
2r*2r-3*2r+6*2r+1=0
Wy multiply elements
4r^2-6r+12r+1=0
We add all the numbers together, and all the variables
4r^2+6r+1=0
a = 4; b = 6; c = +1;
Δ = b2-4ac
Δ = 62-4·4·1
Δ = 20
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}*\sqrt{5}=2\sqrt{5}$$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{5}}{2*4}=\frac{-6-2\sqrt{5}}{8} $$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{5}}{2*4}=\frac{-6+2\sqrt{5}}{8} $
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